I would like to show that for the equation $x^2y+3y+z^3-z-8 = 0$, for every $x$ and $y$, there is a unique $z$ for all $(x,y,z)$ sufficiently near $(1,2,0)$. (I'm working on a problem where I can not directly apply the implicit function theorem, but still have to show that there exists a function $f$ defined locally around $(1,2,0)$ such that $z = f(x,y)$.
It seems fairly obvious to me, since if we arrange the equation as such: $z^3-z = -x^2y-3y +8$, and consider points $(x,y,z)$ within an open ball centered around this point, where $x$ is very close to 1 and $y$ is very close to 2, then this equation becomes $z^3-z=0$, whose roots are $-1,0,1$, and only one of these $z$'s can be within this ball.
I am having a hard time rigorously expressing this, since I'm not sure how to write my intuitive reasoning above. If I let $\epsilon = 1/10$ for example, where $\epsilon$ is the radius of the open ball centered around $(1,2,0)$, I'm left with an inequality $-1/10 < z^3-z < 1/10$, and I'm not sure how to show that all the roots possible are at least $2/10$ away from each other (i.e only root can be in the open ball).
You can do this explicitly because you can solve a cubic.
Recall: The zeros of the depressed cubic $t^3+pt+q$ are given by $$ t_k=2\sqrt{-\frac{p}3}\cos\left(\frac13\cos^{-1}\left(\frac{3q}{2p}\sqrt{\frac{-3}p}\right)-\frac{2\pi k}{3}\right). $$ (Proof: Let $t=2\sqrt{-\frac{p}3}\cos\theta$ and use the triple angle formula for cosine $\cos3\theta=4\cos^3\theta-3\cos\theta$.)
In our case, $z^3-z+(x^2y+3y-8)$ therefore has roots $$ z_k=2\sqrt{\frac13}\cos\left(\frac13\cos^{-1}\left(-\frac32(x^2y+3y-8)\sqrt{3}\right)-\frac{2\pi k}{3}\right), $$ and at $(x,y)=(1,2)$, we have $$ z_k=\frac2{\sqrt3}\cos\left(\frac\pi6-\frac{2\pi k}3\right). $$ Note that $x^2y+3y-8=0$ at $(x,y)=(1,2)$ so we can guarantee $x^2y+3y-8\in(-\frac2{3\sqrt3},\frac2{3\sqrt3})$, hence $z_k$ are real-valued and continuous, for all $(x,y)$ sufficiently close to $(1,2)$). Therefore $(x,y,z_1)$ is sufficiently close to $(1,2,0)$ for $(x,y)$ sufficiently close to $(1,2)$.